一、题目说明
题目198. House Robber，给一列正整数表示每个房间存的金币，不能连续抢2个房间，计算可以得到的最大金币。
二、我的解答
这个题目，我列举了n=1,2,3，...5的情况，没有找到规律。后面看了解答知道了：
dp[i+1]= max(dp[i-2]+nums[i],dp[i-1])
代码如下：
class Solution{
public:
int dfs(vector& nums,int n){
if(n==0) return 0;
if(n==1) return nums[0];
return max(dfs(nums,n-2)+nums[n-1],dfs(nums,n-1));
}
int rob(vector& nums){
return dfs(nums,nums.size());
}
};
遗憾的是，超时：Time Limit Exceeded，通过map，裁剪如下：
class Solution{
public:
int dfs(vector& nums,int n){
if(n==0) return 0;
if(n==1) return nums[0];
if(ump.count(n)>0){
return ump[n];
}
int result = max(dfs(nums,n-2)+nums[n-1],dfs(nums,n-1));
ump[n] = result;
return result;
}
int rob(vector& nums){
return dfs(nums,nums.size());
}
private:
unordered_map ump;
};
Runtime: 4 ms, faster than 58.41% of C++ online submissions for House Robber.
Memory Usage: 9.2 MB, less than 5.66% of C++ online submissions for House Robber.
三、优化措施
用dp方法，代码如下：
class Solution{
public:
int rob(vector& nums){
//dp[i]表示抢第i所房子的最大值，
//dp[i+1]= max(dp[i-2]+nums[i],dp[i-1] )
if(nums.size()<1) return 0;
if(nums.size() == 1) return nums[0];
if(nums.size() == 2) return max(nums[0],nums[1]);
vector dp(nums.size()+1,0);
dp[0] = nums[0];
dp[1] = max(nums[0],nums[1]);
for(int i=2;i 性能如下：
Runtime: 0 ms, faster than 100.00% of C++ online submissions for House Robber.
Memory Usage: 8.6 MB, less than 77.36% of C++ online submissions for House Robber.